Yes, provided p = perimeter, b= breadth and h = height.
6h 8min
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
3h + (2h 20m) + 35m = (3+2)h + (20+35)m = 5h + 55m
Perimeter of a rectangle (P) = 2b+2h = 2*24+2h=114 48+2h=114 2h=66 h=33" Area of a rectangle (A) = b*h = 24*33= 1584 sq. in.
It is: 11h
2h + 2h + 2h = 6h
2H+ + SO42- + Ca2+ + 2I- CaSO4 + 2H+ + 2I-
6h 8min
(t+h)(x+2)
3h-5h + 11 = 17 is ------2h + 11 = 17- 11 -11____________-2h = 6___ ___-2 -2h = -3 (This is the answer.)
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
2Si +2HF-2siF + 2H
Cl2 + 2H = 2HCl
h = 0.538462 14 + 5h + 2h = 5h + 28h 14 + 7h = 33h 14 + 7h - 7h = 33h - 7h 14 = 26h 14/26 = 26/26h 0..538462 = h
Suppose base and height are b and h. 2b+2h=44 (the perimeter includes two bases and two widths) b=2b-11 base is 11 less than two times base Subtract b from both sides: 0=2b-11-b, b-11=0, b=11 2b=22 2b+2h=44, 22+2h=44, 2h=44-22=22, h=11 So the rectangle is a square with side 11.
No because it equals zero.
I2 + 10 hno3 = 2 hio3 + 10 no2 + 4 h2o